Suggestion

Show that the sum to \(n\) terms of the series \[\frac{1}{2.1}-\frac{1}{5.2}-\frac{1}{8.5}-\cdots+ \frac{1}{(3r-1)(4-3r)}\cdots\] is \[\frac{n}{3n-1}.\]

Could we try a proof by induction here?

Or else, could we split \(\dfrac{1}{(3r-1)(4-3r)}\) into partial fractions?

Given that for \[-1<x<+1, \quad \log_e(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-
\quad\text{to infinity,}\] deduce the sum to infinity of \[x+\frac{x^5}{5}+\frac{x^7}{7}+\cdots.\]

How does the sequence continue? Could it be \[x+\frac{x^5}{5}+\frac{x^7}{7}+\frac{x^9}{9}+\frac{x^{11}}{11}+\cdots,\] so that the only odd power missing here is \(\dfrac{x^3}{3}\)?
What would the series for \(\log_e(1-x)\) be? Or the series for \(\log_e(1+x^2)\)?